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DedPeter [7]
3 years ago
11

Find the value of angle x , angle y, and angle z in figure.​

Mathematics
2 answers:
Lynna [10]3 years ago
5 0

\bf \underline{★ Solution-} \\

In the given figure, we are given with two lines which are parallel to each other. We are also given with two lines which forms a triangle and also forms as a transversal lines to the parallel lines. We are also given that the given triangle is an isosceles triangle. So, we can say that the other angle in the triangle also measures 75°.

Now, let's find the value of the ∠x.

We know that the alternate angles in the parallel line always measures the same as the one which is in it's alternate side. So,

\sf \leadsto \angle{x} = {75}^{\circ}

Now, let's find the value of the ∠z.

We know that, all the angles in a triangle always adds up to 180°. In the given triangle, we are given with two angles, so we can easily find the third angle.

\sf \leadsto {75}^{\circ} + {75}^{\circ} + \angle{z} = {180}^{\circ}

\sf \leadsto {150}^{\circ} + \angle{z} = {180}^{\circ}

\sf \leadsto \angle{z} = 180 - 150

\sf \leadsto \angle{z} = {30}^{\circ}

Now, let's find the value of the ∠y.

We know that all the angles that forms a straight line always equals up to 180° (or) the the straight line angle always measures 180°. So, we can find the value of the ∠y by this concept.

\sf \leadsto {75}^{\circ} + {30}^{\circ} + \angle{y} = {180}^{\circ}

\sf \leadsto {105}^{\circ} + \angle{y} = {180}^{\circ}

\sf \dashrightarrow \angle{y} = 180 - 105

\sf \dashrightarrow \angle{y} = {75}^{\circ}

Therefore,

  1. The value of the ∠x is 75°.
  2. The value of the ∠y is 75°.
  3. The value of the ∠z is 30°.

\bf \underline{Hope\: this \: helps!!} \\

SpyIntel [72]3 years ago
4 0

Answer:

x = 75°; y = 75°; z = 30°

Step-by-step explanation:

I am using points A and B as vertices of the triangle.

From the figure, we see that segments BC and AC are congruent. Since they are sides of triangle ABC, that makes their opposite angles, A and B, congruent angles.

m<A = m<B

We are given m<B = 75°,

so m<A = m<B = 75°

In a triangle, the sum of the measures of the interior angles is 180°.

m<A + m<B + z = 180°

Use substitution for angles A and B:

75° + 75° + z = 180°

z = 30°

Lines AB and DE are shown to be parallel. That makes alternate interior angles congruent. Lines AC and BC are transversals to the parallel lines.

m<A = y

y = m<A = 75°

m<B = x

x = m<B = 75°

Answer: x = 75°; y = 75°; z = 30°

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Answer:

No, According to triangle Inequality theorem.

Step-by-step explanation:

Given:

Length given are 4 in., 5 in., 1 in.

We need to check whether with these lengths we can create triangular components.

The Triangle Inequality Theorem states that the sum of any 2 sides of a triangle must be greater than the measure of the third side.

These must be valid for all three sides.

Hence we will check for all three side,

4 in + 5 in > 1 in. (It is a Valid Condition)

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Since 2 condition are valid and 1 condition is not we can say;

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3 years ago
Implicit differentiation Please help
Anvisha [2.4K]

Answer:

y''(-1) =8

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
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  • Left to Right

Equality Properties

<u>Algebra I</u>

  • Factoring

<u>Calculus</u>

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

  • f(x) = cxⁿ
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Product Rule: \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Chain Rule: \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Quotient Rule: \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

-xy - 2y = -4

Rate of change of the tangent line at point (-1, 4)

<u>Step 2: Differentiate Pt. 1</u>

<em>Find 1st Derivative</em>

  1. Implicit Differentiation [Product Rule/Basic Power Rule]:                            -y - xy' - 2y' = 0
  2. [Algebra] Isolate <em>y'</em> terms:                                                                               -xy' - 2y' = y
  3. [Algebra] Factor <em>y'</em>:                                                                                       y'(-x - 2) = y
  4. [Algebra] Isolate <em>y'</em>:                                                                                         y' = \frac{y}{-x-2}
  5. [Algebra] Rewrite:                                                                                           y' = \frac{-y}{x+2}

<u>Step 3: Find </u><em><u>y</u></em>

  1. Define equation:                    -xy - 2y = -4
  2. Factor <em>y</em>:                                 y(-x - 2) = -4
  3. Isolate <em>y</em>:                                 y = \frac{-4}{-x-2}
  4. Simplify:                                 y = \frac{4}{x+2}

<u>Step 4: Rewrite 1st Derivative</u>

  1. [Algebra] Substitute in <em>y</em>:                                                                               y' = \frac{-\frac{4}{x+2} }{x+2}
  2. [Algebra] Simplify:                                                                                         y' = \frac{-4}{(x+2)^2}

<u>Step 5: Differentiate Pt. 2</u>

<em>Find 2nd Derivative</em>

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